Question: $ E = \left[\begin{array}{r}1 \\ 4 \\ 1\end{array}\right]$ $ C = \left[\begin{array}{rr}-2 & 2\end{array}\right]$ What is $ E C$ ?
Solution: Because $ E$ has dimensions $(3\times1)$ and $ C$ has dimensions $(1\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ E C = \left[\begin{array}{r}{1} \\ {4} \\ \color{gray}{1}\end{array}\right] \left[\begin{array}{rr}{-2} & \color{#DF0030}{2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ E$ , with the corresponding elements in column $j$ of the second matrix, $ C$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ E$ with the first element in ${\text{column }1}$ of $ C$ , then multiply the second element in ${\text{row }1}$ of $ E$ with the second element in ${\text{column }1}$ of $ C$ , and so on. Add the products together. $ \left[\begin{array}{rr}{1}\cdot{-2} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ E$ with the corresponding elements in ${\text{column }1}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{-2} & ? \\ {4}\cdot{-2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ E$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ C$ and add the products together. $ \left[\begin{array}{rr}{1}\cdot{-2} & {1}\cdot\color{#DF0030}{2} \\ {4}\cdot{-2} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{1}\cdot{-2} & {1}\cdot\color{#DF0030}{2} \\ {4}\cdot{-2} & {4}\cdot\color{#DF0030}{2} \\ \color{gray}{1}\cdot{-2} & \color{gray}{1}\cdot\color{#DF0030}{2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-2 & 2 \\ -8 & 8 \\ -2 & 2\end{array}\right] $